Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.Collections;


public class Funky {
public static void main(String[] args) throws Exception {
        StreamTokenizer st = new StreamTokenizer(new BufferedReader(
                new InputStreamReader(System.in)));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        
        ArrayList<Integer> tri = new ArrayList<Integer>();
        
        for (long k=1; k*(k+1)/2L<=1000000000; k++)
            tri.add((int)(k*(k+1)/2L));
        
        st.nextToken();
        int n = (int) st.nval;
        
        boolean found = false;
        
        for (int i=0; i<tri.size(); i++)
        {
            int first = tri.get(i);
            int req = n - first;
            
            if (req < 0)
                continue;
            
            if (Collections.binarySearch(tri, req) >= 0)
                found = true;
        }
        
        if (found)
            pw.write("YES\n");
        else
            pw.write("NO\n");
        
        
        pw.flush();
        pw.close();
    }
}